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B Magnitude of gravitational time dilation

  1. Apr 21, 2016 #1
    Hello,
    I am wondering how is the magnitude of the time dilation in a gravity field related to the time dilation we know from special relativity. How does the dilation caused by just being in the field compare to the dilation you'd have from your velocity if you fell into the gravity field from infinity?
     
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  3. Apr 21, 2016 #2

    stevendaryl

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    Near the Earth, using Scharzschild coordinates, for motion in the radial direction, the time dilation formula is given by:

    [itex]d\tau = \sqrt{Q - \frac{\frac{v^2}{c^2}}{Q}} dt[/itex]

    where [itex]Q[/itex] is [itex]1 - \frac{2GM}{c^2 r}[/itex], and where [itex]M[/itex] is the mass of the Earth, and where [itex]t[/itex] is coordinate time (time as measured by a clock far from the Earth).

    This might be easier to understand in the nonrelativistic, weak-gravity limit. In this limit, this expression becomes approximately:

    [itex]\frac{d\tau}{dt} \approx 1 - \frac{GM}{c^2 r} - \frac{1}{2} \frac{v^2}{c^2}[/itex]

    So you can sort of think of this as the first expression being a gravity-dependent time dilation, which says that clocks with greater gravitational potential energy run faster, plus a velocity-dependent time dilation, which says that clocks with greater speed run slower. For a clock on a satellite orbiting the Earth, these two effects work in opposite directions, because its gravitational potential energy is higher, making it run faster than clocks on the ground, but its velocity is also higher, making it run slower than clocks on the ground. You have to look at the details of the orbit to see which effect dominates. For a circular orbit,

    [itex]v^2 = \frac{GM}{r}[/itex]

    So the "velocity-dependent" term, [itex]\frac{1}{2} \frac{v^2}{c^2}[/itex], is half of the "gravity-dependent" term.
     
  4. Apr 21, 2016 #3

    stevendaryl

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    You asked about the case of falling from infinity. In that case, [itex]v^2 = \frac{2 GM}{c^2 r}[/itex] (as computed using Newtonian physics, which is pretty accurate near the Earth). So the velocity-dependent time dilation is approximately equal to the gravity-dependent time dilation.
     
  5. Apr 21, 2016 #4
    Thanks, this is what I suspected. Though I am surprised you say "approximately equal". I see your point about using an approximation for the velocity in the above post, but isn't there a similar approximation for the potential energy term in your first post as well? Would the two types of dilations be equal even without approximations?
     
  6. Apr 21, 2016 #5
    To clarify I am not wondering about weak-gravity limits and approximations here. The original reason I asked was the example of "Miller's planet" orbiting a black hole in the "Interstellar" movie. It is said to have a dilation factor turning an hour into 7 years, or over 60000. So I wondered if the kinetic energy of the velocity needed to get the same dilation factor is a good estimate of the potential energy change needed to enter or exit that orbit.
     
  7. Apr 21, 2016 #6
    To get to a place where time ticks 100 times faster compared to your current place, you need fuel 100 times your own mass.

    When you accelerate yourself using fuel 100 times your own mass, you reach speed that corresponds to relativistic factor 100. (0.99995 c)

    Do I need to give reasons? Well climbing the uphill multiplies your energy by factor 0.01, so you need to multiply your energy by factor 100, in order to arrive to the destination with your original energy. (I'm talking about your energy according to yourself here)
     
    Last edited: Apr 21, 2016
  8. Apr 21, 2016 #7

    PeterDonis

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    This is somewhat confused. Your "energy according to yourself" is always the same--it's just your rest mass. The "energy" you are talking about that has to be multiplied by 100 is your "energy at infinity", which is, heuristically, the energy an observer at infinity would attribute to you. If you are so close to a black hole horizon that your energy at infinity is only 0.01 of your rest mass, the observer at infinity would say you are very tightly bound to the black hole--so much so that you had to emit energy to infinity equal to 99 percent of your rest mass in order to get into that tightly bound state. (The 99 percent of your rest mass is also sometimes referred to as "gravitational binding energy".)

    When you climb back uphill, you have to therefore increase your energy at infinity back to its original value at infinity--which is just your rest mass. That means you have to absorb energy equal to 99 percent of your rest mass--or, to put it another way, that amount of work has to be done on you to raise you uphill. This process does not "multiply your energy by 0.01"--it multiplies your energy at infinity by 100. If the work is done on you by means of a rocket with fuel that starts out with you, then yes, that fuel has to have rest mass 99 times yours (assuming 100% conversion of fuel mass to energy and perfect collimation of the rocket exhaust). But from the standpoint of energy at infinity, that means you + your fuel are sitting way down in the gravity well, with total energy at infinity equal to your rest mass; then all of the energy at infinity in the fuel is transferred to you, raising you up to infinity, where you have energy at infinity equal to your rest mass. So the total energy at infinity of the system of you + fuel did not change at all. (Note that we're assuming the fuel has zero energy at infinity when the process is complete, which is not really possible, but we can ignore that here.)
     
  9. Mar 21, 2017 #8
    I have seen this formula without the second Q in the denominator so I wonder why do you have the second Q in the denominator but the Wikipedia page for this time dilation expression does not have the Q in the denominator like yours here? The other equation is simply [itex]d\tau = \sqrt{Q - {\frac{v^2}{c^2}}{}} dt[/itex]. Which one applies to which situations if both are true?
     
  10. Mar 21, 2017 #9

    pervect

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    The case described in "Interstellar" isn't the same one described by stevendaryl, it doesn't use the Schwarzschild metric. The case described in "Interstellar" uses the Kerr metric of a rotating black hole - one rotating nearly as fast as possible, a sort of rotating black hole called "extremal". The physical situation is different - the "gravity" of such an extremal black hole is different from the "gravity" of a non-rotating Schwarzschild black hole.

    I believe Kip Thorne did the calculations for the extremal black hole in "The Science of Interstellar" https://www.amazon.com/Science-Interstellar-Kip-Thorne/dp/0393351378, though I don't have a copy. I recall a few posters here doing the relevant calculation, as well. If you're not familiar with the details of GR, though, the details of the calculation may not make a lot of sense. I believe I heard that the movie director wanted such an example of a large time dilation factor. Kip Thorne, the movie's science advisor and author of the previously mentioned book (among other books on GR), hand-crafted the system to match the desires of the movie's producer. I believe that the resulting system was described by Thorne as "unlikely" and not-very-stable in the long term, but Thorne deemed it physically possible.

    [add]Let me add that while Stevedaryl didn't analyze the case you were thinking of, the case he analyzed is considerably easier to calculate and understand. So, for someone trying to understand gravitational time dilation, Thorne's case is probably not the best place to start, the simpler case that's not from "interstellar" that Stevedaryl analyzed is a better place to start.
     
    Last edited by a moderator: May 8, 2017
  11. Mar 21, 2017 #10

    PeterDonis

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    The second term under the square root might or might not have the ##Q## in the denominator, depending on the direction of ##v##. If ##v## is purely tangential, there is no ##Q## there; if it's purely radial, as you note, there is a ##Q## there; if it's a mixture, you need two terms, one with ##Q## and one without.

    (The above also assumes that ##v## is coordinate velocity, which will not be the same as the velocity actually measured by a static observer at that ##r## coordinate.)
     
  12. Mar 21, 2017 #11

    PeterDonis

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    See my post just now.
     
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